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Ace2
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May 14th, 2023 at 7:55:02 AM permalink

[[{(38/18)^(5+1) - 1} / {(38/18) - 1}] - 1] / 2 =~ 38.9 spins
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May 15th, 2023 at 5:31:17 AM permalink
Quote: ThatDonGuy

Quote: Wizard

What is the expected number of spins in double-zero roulette to see five reds or five blacks in a row? The previous spin landed on 0.
link to original post



Let E(n) be the expected number after having n consecutive of a color; E(5) = 0
Let p be the probability of getting a particular color, and r = 1 - 2p = the probability that it comes up green

E(4) = 1 + p E(1) + r E(0)

E(3) = 1 + p E(1) + r E(0) + p E(4)
= 1 + p E(1) + r E(0) + p (1 + p E(1) + r E(0))
= 1 + p + (p + p^2) E(1) + (r + pr) E(0)

E(2) = 1 + p E(1) + r E(0) + p E(3)
= 1 + p E(1) + r E(0) + p (1 + p + (p + p^2) E(1) + (r + pr) E(0))
= (1 + p + p^2) + p (1 + p + p^2) E(1) + r (1 + p + p^2) E(0)

E(1) = 1 + p E(1) + r E(0) + p E(3)
= (1 + p + p^2 + p^3) + p (1 + p + p^2 + p^3) E(1) + r (1 + p + p^2 + p^3) E(0)
(1 - p - p^2 - p^3 - p^4) E(1) = (1 + p + p^2 + p^3) + (1 - 2p) (1 + p + p^2 + p^3) E(0)
E(1) = (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4) + ((1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0)

E(0) = 1 + (1 - 2p) E(0) + 2p E(1)
E(0) = 1 / 2p + E(1)
E(0) = 1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4) + ((1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0)
(1 - (1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0) = 1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)
E(0) = (1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) x (1 - p - p^2 - p^3 - p^4) / p^4
E(0) = (1 - p - p^2 - p^3 - p^4) / (2 p^5) + (1 + p + p^2 + p^3) / p^4
E(0) = (1 - p - p^2 - p^3 - p^4) / (2 p^5) + (2p + 2p^2 + 2p^3 + 2p^4) / (2p^5)
= (1 + p + p^2 + p^3 + p^4) / (2p^5)
= (1 - p^5) / (2 p^5 (1 - p))

For p = 9/19, this is 4,592,395 / 118,098, or about 38.8863


link to original post



I agree! That one is beer worthy.

Eliot's approximation was correct to the four significant digits given. Of course, an exact expression is required for the beer.

Ace is also correct and extra credit for the formula for n spins.
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teliot
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May 15th, 2023 at 3:05:33 PM permalink
What I finds odd about this problem is this related problem: what are the chances that starting with the present spin you get a sequence of exactly 5 reds or 5 blacks?

The answer is, of course, (36/38)*(18/38)^4*(20/38) = 0.025103 or 1-in-39.836. I find it odd that this is even close to the other question Mike asked (the expected number of spins to get a sequence Red or Black of length 5). I was almost going to go with 39.836 as my answer.
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Ace2
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May 15th, 2023 at 3:50:35 PM permalink
Quote: teliot

What I finds odd about this problem is this related problem: what are the chances that starting with the present spin you get a sequence of exactly 5 reds or 5 blacks?

The answer is, of course, (36/38)*(18/38)^4*(20/38) = 0.025103 or 1-in-39.836. I find it odd that this is even close to the other question Mike asked (the expected number of spins to get a sequence Red or Black of length 5). I was almost going to go with 39.836 as my answer.
link to original post

With the last spin being zero, the chance of getting five black or five red in the next five spins is (18/38)^5 * 2 or about 1 in 21. Not close to the answer

Look at it this way. The chance of you getting two consecutive heads in the next two rolls is 1 in 4, but the expected waiting time to see two consecutive heads is six. Red/black is not much different from heads/tails…would be identical without those dang two zeroes on the wheel
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teliot
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May 15th, 2023 at 4:11:10 PM permalink
Of course, but I was giving exactly 5 in a row, not 5 or more.

I'm just saying that two unrelated questions about five in a row have surprisingly close answers.
Last edited by: teliot on May 15, 2023
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May 25th, 2023 at 12:17:58 PM permalink
A cake is divided among 100 mathematicians, numbered 1 to 100. It is divided up as follows an in order:

Mathematician 1 gets 1% of the cake.
Mathematician 2 gets 2% of the remaining cake.
Mathematician 3 gets 3% of the remaining cake.
Mathematician 4 gets 4% of the remaining cake.

This continues until the last mathematician gets 100% of what is left.

Which mathematician gets the most cake?

No calculators allowed!

For full credit, I need an explanation why your answer is right.

I also hate to keep saying this, but please don't just throw out a link to a video that solves the problem *ahem*. Let's have some fun and try to learn something.
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charliepatrick
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May 25th, 2023 at 12:43:46 PM permalink
Person N gets N/100 of what's left. In which case person (N+1) gets (N+1)/100*(100-N)/100. This means 100N=100N+100-N2-N which means N2+N-100=0. Using N=-b2+/-SQRT(etc) gives -1+-SQRT(1+400)/2 which is about 10.
When N=10 the 10th person gets 10% and the 11th person get 11% of 90% = 9.9%. Similarly the 9th person gets 9% and the 10th person gets 10% of 91% = 9.1%. Confirming the 10th person gets the most.
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May 25th, 2023 at 1:24:42 PM permalink
Quote: charliepatrick

Person N gets N/100 of what's left. In which case person (N+1) gets (N+1)/100*(100-N)/100. This means 100N=100N+100-N2-N which means N2+N-100=0. Using N=-b2+/-SQRT(etc) gives -1+-SQRT(1+400)/2 which is about 10.
When N=10 the 10th person gets 10% and the 11th person get 11% of 90% = 9.9%. Similarly the 9th person gets 9% and the 10th person gets 10% of 91% = 9.1%. Confirming the 10th person gets the most.

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Agreed. Full credit for a similar solution as mine.
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lilredrooster
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May 26th, 2023 at 2:20:00 AM permalink
.

𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵___________________that is part of the lyrics from the Beatles song "Revolution"


the 9 times mystery:


at least it's a mystery to me..............I'm sure some here can explain it - an easy puzzle I'm sure for some - but not for me
I haven't got a clue
nothing similar happens with any other digits that I know about


here we go:


9*5=45.......................add up the digits - 4+5=9

9*8=72.......................add up the digits - 7+2=9

9*7=63.......................add up the digits - 6+3=9

9*17=153...................add up the digits - 1+5+3 =9

9*77=693...................add up the digits - 6+9+3 and get 18............then add up those digits..............1+8=9

9*4,796=43,164..........add up the digits 4+3+1+6+4 and get 18......then add up those digits...............1+8=9

9*104,675 = 942,075.........add up the digits..........9+4+2+0+7+5 and get 27...........then add up those digits..............2+7=9

9*5,327,894 = 47,951,046...........add up the digits .........4+7+9+5+1+0+4+6 and get 36............then add up those digits.................3+6=9



𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵

.
Please don't feed the trolls
charliepatrick
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May 26th, 2023 at 2:51:09 AM permalink
You are correct that the sum of the digits of a number divisible by 9 will, assuming you keep going, always land up as 9. The logic can be thought as follows.

For two digits numbers the number can be expressed as 10a+b, where as the total of the digits is a+b. This if a+b=9, then 9a+a+b is also a multiple of 9, and this is 10a+b.

For three digits use 100a+10b+c and observe that 99a+9b+(a+b+c) is also divisible by 9 if (a+b+c) is. If a+b+c is greater than 9 then repeat the process, and note the same logic applies for the number formed by a+b+c.

For larger numbers you can see that each power of 10 can be considered as (99..99+1) so the number breaks down to 9*something+(a+b+.....+z). So you can see the pattern applies for all numbers.

btw the same applies to numbers divisible by 3, the total wil always reduce to 3 6 or 9.
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May 26th, 2023 at 3:15:49 PM permalink
It takes Alice and Bill 3 hours to paint a house.
It takes Alice and Cindy 4 hours to paint a house.
It takes Bill and Cindy 5 hours to paint a house.

How long does it take if they all paint?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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May 26th, 2023 at 3:46:09 PM permalink
Quote: Wizard

It takes Alice and Bill 3 hours to paint a house.
It takes Alice and Cindy 4 hours to paint a house.
It takes Bill and Cindy 5 hours to paint a house.

How long does it take if they all paint?
link to original post



Let A, B, and C be the amount of the house Alice, Bill, and Cindy can paint in an hour
A + B = 1/3
A + C = 1/4
B + C = 1/5
2 (A + B + C) = 1/3 + 1/4 + 1/5 = 47/60
A + B + C = 47/120
It takes all three 120/47, or 2 26/47, hours

charliepatrick
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May 26th, 2023 at 3:51:41 PM permalink
Quote: Wizard

It takes Alice and Bill 3 hours to paint a house.
It takes Alice and Cindy 4 hours to paint a house.
It takes Bill and Cindy 5 hours to paint a house.

How long does it take if they all paint?
link to original post

Let A be how many rooms A can paint in an hour, etc.
3A+3B=4A+4C=5B+5C=size of house
3A=2B+5C
4A=5B+C
8B+20C=15B+3C
17C=7B
Let B=17 C=7 gives A=23
This has A+B=40, A+C=30, B+C=24
So if this is how many rooms they can each paint, then the house has 120 rooms.
So the time for all of them to paint house is 120/47 = 2h33:11.489s
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May 26th, 2023 at 5:38:50 PM permalink
Quote: ThatDonGuy


Let A, B, and C be the amount of the house Alice, Bill, and Cindy can paint in an hour
A + B = 1/3
A + C = 1/4
B + C = 1/5
2 (A + B + C) = 1/3 + 1/4 + 1/5 = 47/60
A + B + C = 47/120
It takes all three 120/47, or 2 26/47, hours


link to original post



I agree! Charlie was right too, but five minutes behind you.
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MichaelBluejay
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May 28th, 2023 at 12:58:23 PM permalink
How about a variant?

It takes Alice 3 hours to paint a house.
It takes Bill 4 hours to paint a house.
It takes Cindy 5 hours to paint a house.

How long does it take if they all paint?

Also, I hope I can hire any of them at a normal hourly rate for a painter, because their throughput is phenomenal.
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MichaelBluejay
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May 28th, 2023 at 1:05:15 PM permalink
This is from my local weekly newspaper's math column. I didn't get this one, because I'm not good with spatial concepts.

Flip the triangle upside-down by moving only 3 balls.

⬤⬤⬤⬤
⬤⬤⬤
⬤⬤


Move the top-left and top-right balls down to the row of 2, making it a row of 4.
Then move the single ball at the bottom to the very top.
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ThatDonGuy
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May 28th, 2023 at 2:27:46 PM permalink
Quote: MichaelBluejay

How about a variant?

It takes Alice 3 hours to paint a house.
It takes Bill 4 hours to paint a house.
It takes Cindy 5 hours to paint a house.

How long does it take if they all paint?

Also, I hope I can hire any of them at a normal hourly rate for a painter, because their throughput is phenomenal.
link to original post



In one hour, Alice paints 1/3 of a house, Bill paints 1/4 of a house, and Cindy paints 1/5 of a house
Togethger, they paint 1/3 + 1/4 + 1/5 = 47 / 60 of a house in an hour.
They can paint one house in 60 / 47 hours

aceside
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May 29th, 2023 at 5:31:29 AM permalink
Quote: MichaelBluejay

This is from my local weekly newspaper's math column. I didn't get this one, because I'm not good with spatial concepts.

Flip the triangle upside-down by moving only 3 balls.

⬤⬤⬤⬤
⬤⬤⬤
⬤⬤


Move the top-left and top-right balls down to the row of 2, making it a row of 4.
Then move the single ball at the bottom to the very top.

link to original post


This solution is a little confusing. To be clearer, I would say, move the top-left and top-right balls 2 rows down making it the bottom row.
teliot
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May 29th, 2023 at 6:20:21 AM permalink
Not clear, but I thought it was a right triangle, so my solution is:

Move top row, two on left to 3rd row left.
Move bottom to new top row on right.

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MichaelBluejay
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May 29th, 2023 at 10:26:32 AM permalink
Quote: aceside

This solution is a little confusing. To be clearer, I would say, move the top-left and top-right balls 2 rows down making it the bottom row.

That wouldn't be clearer, because it would be wrong. The two balls move down to the penultimate row, not the bottom row.

Also, you've revealed part of the solution outside the spoiler, but I guess the ball is out of the bag now.
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May 30th, 2023 at 6:15:11 AM permalink
Quote: MichaelBluejay

This is from my local weekly newspaper's math column. I didn't get this one, because I'm not good with spatial concepts.

Flip the triangle upside-down by moving only 3 balls.

⬤⬤⬤⬤
⬤⬤⬤
⬤⬤


link to original post



Here is another way of putting the solution.



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MichaelBluejay
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May 30th, 2023 at 12:15:51 PM permalink
Quote: Wizard

Here is another way of putting the solution.

The color-coding is exceptionally clear.
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charliepatrick
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June 3rd, 2023 at 6:22:51 AM permalink
Seats in a Cafe

I was sitting in a local cafe earlier this week and noticed that all the chairs were either red, black or white. Also each table had four seats. It got me thinking if all the tables had been circular (i.e. a seating arrangement RRBW is the same as RBWR etc.) and every table had four chairs with a different arrangement from any other table, what would be the largest possibe cafe size.

In other words how many possible ways are there to arrange four chairs, each being either Red, Black or White, around a circular table.

I think it's fair to say that the answer is bigger than the number of tables my local cafe actually has!!
ThatDonGuy
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June 3rd, 2023 at 7:06:17 AM permalink
Quote: charliepatrick

Seats in a Cafe

I was sitting in a local cafe earlier this week and noticed that all the chairs were either red, black or white. Also each table had four seats. It got me thinking if all the tables had been circular (i.e. a seating arrangement RRBW is the same as RBWR etc.) and every table had four chairs with a different arrangement from any other table, what would be the largest possibe cafe size.

In other words how many possible ways are there to arrange four chairs, each being either Red, Black or White, around a circular table.

I think it's fair to say that the answer is bigger than the number of tables my local cafe actually has!!
link to original post



The possible sets of chairs:
4,0,0: 3 (colors) x 1 (permutation) = 3
RRRR, WWWW, BBBB
3,1,0: 3 (colors for the 3) x 2 (colors for the 1) x 1 (permutation) = 6
RRRW, RRRB
BBBR, BBBW
WWWR, WWWB
2,2,0: 3 (pairs of colors) x 2 (permutations) = 6
RRBB, RBRB
RRWW, RWRW
BBWW, BWBW
2,1,1: 3 (colors for the pair) x 3 (permutations) = 9
RRBW, RRWB, RWRB
BBWR, BBRW, BWBR
WWRB, WWBR, WBWR
Total = 3 + 6 + 6 + 9 = 24 tables

charliepatrick
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June 3rd, 2023 at 7:11:58 AM permalink
^ Yes that's how I did it.
aceside
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June 3rd, 2023 at 7:24:13 AM permalink
Deleted
aceside
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June 3rd, 2023 at 7:26:12 AM permalink
I used ChatGPT to find out, by Burnside’s lemma, there are (81+9+3+3)/4=24 ways.
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June 5th, 2023 at 5:34:35 PM permalink


Suppose you have a partially full bottle of pancake syrup in an unusually shaped bottle.

The bottle is 12" high.

When the cap it at the top, the syrup makes it to 7" point.

When the bottle is turned upside down, the syrup makes it to the 9" point.

What ratio of the bottle is full?
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MichaelBluejay
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June 5th, 2023 at 5:44:13 PM permalink
Um, 2/3? Is it as simple as taking the average of 7 and 9, and dividing by 12?
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June 5th, 2023 at 7:22:10 PM permalink
Quote: MichaelBluejay

Um, 2/3? Is it as simple as taking the average of 7 and 9, and dividing by 12?

link to original post



I started to write a post why your answer was wrong, but in the process I think I find that there isn't enough information given to solve it.

I request we change the shape of the bottle to a wine bottle, where the vast majority is shaped like a cylinder.

All due apologies for the bad question.
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June 5th, 2023 at 7:38:41 PM permalink


My guess assuming the bottom 7 inches are all cylindrical....

Turn it upside down and the first 5" is what was above the previous 7" of wine.
The next 4" are what would have been the top 4" of the previous 7" of wine.

So the first 5" (when inverted) must be the same volume as the bottom 3" when right side up.

So if just a cylinder, it would be 3" + 4" (original 7") + 3" (the equivalent of the 5" inverted)

So original is 7/10 = 70%

But then again I don't like wine, so what do I know.
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June 5th, 2023 at 7:51:27 PM permalink
Quote: chevy



My guess assuming the bottom 7 inches are all cylindrical....

Turn it upside down and the first 5" is what was above the previous 7" of wine.
The next 4" are what would have been the top 4" of the previous 7" of wine.

So the first 5" (when inverted) must be the same volume as the bottom 3" when right side up.

So if just a cylinder, it would be 3" + 4" (original 7") + 3" (the equivalent of the 5" inverted)

So original is 7/10 = 70%

But then again I don't like wine, so what do I know.

link to original post



I agree!
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ThatDonGuy
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June 6th, 2023 at 8:41:36 AM permalink
Quote: Wizard

Suppose you have a partially full bottle of pancake syrup in an unusually shaped bottle.

The bottle is 12" high.

When the cap it at the top, the syrup makes it to 7" point.

When the bottle is turned upside down, the syrup makes it to the 9" point.

What ratio of the bottle is full?
link to original post



Assume the bottle is in the shape of two cylinders, one of height 7 and base area 1, and the other of height 5 and base area B.

When the base 1 cylinder is on the bottom, the liquid fills it, but there is no excess that goes into the base B cylinder
The volume = 7 x 1 = 7

When the base B cylinder is on the bottom, the liquid fills it, and there is an excess that partially fills the base 1 cylinder to a height of 4
The volume = B x 5 + 1 x 4 = 4 + 5B

Since the volume of the liquid is the same in both cases, 4 + 5B = 7, or B = 3/5
The total volume of the bottle = 7 x 1 + 5 x 3/5 = 10
The ratio that is full in this case is 7 / 10

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June 6th, 2023 at 8:58:30 AM permalink
Away all boats! On this 79th anniversary of the day that Eisenhower knocked the Allied occupation of Rome off of the newspaper headlines, not to mention the 79th anniversary of the combat debut of the Screaming Eagles (the 101st Airborne Division), here's an "easy" problem:

1. Express each of the integers from 1 to 28 in the form:

A # B # C # D

where A, B, C, D are some permutation of 1, 2, 3, 4, and each # is either the addition, subtraction, or multiplication operators, but not division (they do not have to be the same operator - e.g. 1 can be 2 x 3 - 4 - 1); you may use parentheses, but there must be an operator between each pair of adjacent numbers rather than using "implied multiplication."
You also cannot use exponentiation (e.g. 23) or concatenating two digits (e.g. using 2 and 4 to make 42).

2. Express each of the integers from 1 to 15 the same way, but without parentheses, and using standard precedence.
Oh, wait; you can't - but you can do it for all but one of them.
Which one? You tell me - then express that one without parentheses if you are allowed to include the division operator.
charliepatrick
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June 6th, 2023 at 9:57:21 AM permalink
7=3x4/2+1.
charliepatrick
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June 6th, 2023 at 11:04:18 AM permalink
1
3x2-4-1
2
3+2+1-4
3
3x2-4+1
4
4x2-3-1
5
4x1+3-2
6
4x2-3+1
7
3x(4-2)+1
7
3x4/2-1
8
4+3+2-1
9
4+3+2x1
10
4+3+2+1
11
4x3-2+1`
12
4x2+3+1
13
4x3+2-1
14
4x3+2x1
15
4x3+2+1
16
4x(3+2-1)
17
(4+2)x3-1
18
(4+2)x3x1
19
(4+2)x3+1
20
(3+2)x4x1
21
(4+2+1)x3
22
(4x3-1)x2
23
4x3x2-1
24
4x3x2x1
25
4x3x2+1
26
(4x3+1)x2
27
3x(4x2+1)
28
4x(3x2+1)
Please accept any typos!
Wizard
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June 6th, 2023 at 5:18:33 PM permalink


A rubber band is wrapped around two circles. The large circle has radius 5. The small circle has radius 2.

What is the distance of the rubber band?

For purposes of discussion, feel free to use...




θ is an angle.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
chevy
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June 6th, 2023 at 6:06:20 PM permalink
Quote: Wizard



A rubber band is wrapped around two circles. The large circle has radius 5. The small circle has radius 2.

What is the distance of the rubber band?

For purposes of discussion, feel free to use...




θ is an angle.

link to original post







mine based on your original drawing and labels.....


d = b = 2
a+b = 5+2 = 7
d+e = 2+e = 5 .... so e = 3

So c^2 = 7^2 - 3^2 = 40.....c=sqrt(40)=2sqrt(10)

There are 2 "c" legs for rubber band

Angle t = arccos(3/7) (in radians)
So angle of rubber stretched on big circle = (2pi - 2t)
So length of rubber band around big circle = (2pi-2t)*5 = 10pi - 10arccos(3/7)

small angle in Wiz's triangle is arcsin(3/7) (in radians)
So angle of rubber stretched on small circle = 2pi - 2*(pi/2 + arcsin(3/7)) = pi-2arcsin(3/7)
So length of rubber band around small circle = (pi-2arcsin(3/7))*2

total = 4sqrt(10) + (10pi-10arccos(3/7)) + (2pi - 4arcsin(3/7))

= 4 sqrt(10) + 12pi - 10arccos(3/7) - 4arcsin(3/7)

(+/- typos....and if you need something without arches or arcsin....I wave the red flag of submission)

Wizard
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June 6th, 2023 at 7:35:50 PM permalink
Quote: chevy





mine based on your original drawing and labels.....


d = b = 2
a+b = 5+2 = 7
d+e = 2+e = 5 .... so e = 3

So c^2 = 7^2 - 3^2 = 40.....c=sqrt(40)=2sqrt(10)

There are 2 "c" legs for rubber band

Angle t = arccos(3/7) (in radians)
So angle of rubber stretched on big circle = (2pi - 2t)
So length of rubber band around big circle = (2pi-2t)*5 = 10pi - 10arccos(3/7)

small angle in Wiz's triangle is arcsin(3/7) (in radians)
So angle of rubber stretched on small circle = 2pi - 2*(pi/2 + arcsin(3/7)) = pi-2arcsin(3/7)
So length of rubber band around small circle = (pi-2arcsin(3/7))*2

total = 4sqrt(10) + (10pi-10arccos(3/7)) + (2pi - 4arcsin(3/7))

= 4 sqrt(10) + 12pi - 10arccos(3/7) - 4arcsin(3/7)

(+/- typos....and if you need something without arches or arcsin....I wave the red flag of submission)


link to original post



I agree! I couldn't express is more simply than that, so full credit!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Wizard
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June 7th, 2023 at 4:45:52 PM permalink
This one is probably not "easy," so all due apologies for putting it here.

Consider a game of rock-paper-scissors where:

  • If rock beats scissors, scissors pays rock $1
  • If scissors beats paper, paper pays scissors $2
  • If paper beats rock, rock pays paper $3
  • No money changes hands on a tie


Assume two logicians are playing. What is the optimal strategy for each?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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June 7th, 2023 at 6:50:43 PM permalink
Quote: Wizard

This one is probably not "easy," so all due apologies for putting it here.

Consider a game of rock-paper-scissors where:

  • If rock beats scissors, scissors pays rock $1
  • If scissors beats paper, paper pays scissors $2
  • If paper beats rock, rock pays paper $3
  • No money changes hands on a tie


Assume two logicians are playing. What is the optimal strategy for each?
link to original post



Let p be the probability of playing paper, and s the probability of playing scissors; (1 - p - s) is the probability of playing rock

Expected values of the strategy, if the opponent plays:
Rock: 3p - s
Paper: 2s - 3(1 - p - s) = 5s + 3p - 3
Scissors: (1 - p - s) - 2p = 1 - s - 3p

Assuming an equilibrium (i.e. all EVs = 0):
5s + 3p - 3 = 3p - s -> s = 1/2
1 - s - 3p = 3p - s -> p = 1/6
Check:
Rock = 3p - s = 0
Paper = 5s + 3p - 3 = 0
Scissors = 1 - s - 3p = 0

Each player rolls a 6-sided die, and plays scissors on 1-3, rock on 4-5, and paper on 6

Wizard
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June 7th, 2023 at 7:40:34 PM permalink
Quote: ThatDonGuy


Let p be the probability of playing paper, and s the probability of playing scissors; (1 - p - s) is the probability of playing rock

Expected values of the strategy, if the opponent plays:
Rock: 3p - s
Paper: 2s - 3(1 - p - s) = 5s + 3p - 3
Scissors: (1 - p - s) - 2p = 1 - s - 3p

Assuming an equilibrium (i.e. all EVs = 0):
5s + 3p - 3 = 3p - s -> s = 1/2
1 - s - 3p = 3p - s -> p = 1/6
Check:
Rock = 3p - s = 0
Paper = 5s + 3p - 3 = 0
Scissors = 1 - s - 3p = 0

Each player rolls a 6-sided die, and plays scissors on 1-3, rock on 4-5, and paper on 6


link to original post



I agree! Not only the correct answer, but the solution was very simply and elegantly stated.

That one is beer worthy!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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June 8th, 2023 at 5:57:27 AM permalink
I was a little surprised that this even had a solution, so I tried to see if there was a general solution, where Rock receives R if it wins, Paper receives P, and Scissors receives S, and in this case, there is:


The expected value based on each of the opponent's plays is:
Rock: q P - (1 - p - q) R = p R + q (P + R) - R
Paper: (1 - p - q) S - p P = S - p (P + S) - q S
Scissors: p R - q S
Assuming a Nash Equilibrium, EV(Rock) = EV(Scissors) when:
p R + q (P + R) - R = p R - q S
q = R / (R + P + S)
EV(Paper) = EV(Scissors) when:
S - p (P + S) - q S = p R - q S
p = S / (R + P + S)
1 - q - p = P / (R + P + S)

Check:
Rock: p R + q (P + R) - R = RS / (R + P + S) + R (P + R) / (R + P + S) - R = 0
Paper: S - p (P + S) - q S = S - (P + S) S / (R + P + S) - RS / (R + P + S) = 0
Scissors: RS / (R + P + S) - SR / (R + P + S) = 0

Play Rock, Paper, Scissors in the proportion S, R, P
i.e. play each in proportion to what the move that this move beats would collect if it was the winning move

aceside
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June 9th, 2023 at 6:41:13 AM permalink
I recently played a 6-deck automatic-shuffle blackjack game. Everything was normal except one thing, that is, 3 of the 6 decks of cards had a red back and the remaining 3 decks of cards had a blue back. Can a player gain some EV edge from this game by taking into account the card back color information?
ThatDonGuy
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June 9th, 2023 at 4:38:14 PM permalink
Quote: aceside

I recently played a 6-deck automatic-shuffle blackjack game. Everything was normal except one thing, that is, 3 of the 6 decks of cards had a red back and the remaining 3 decks of cards had a blue back. Can a player gain some EV edge from this game by taking into account the card back color information?
link to original post


If it wasn't automatic shuffle, I would say yes - for example, if you have seen all 12 blue-backed Aces come out and the next card is blue, then you know it's not an Ace - although I doubt you could get much better information than by straight card counting, especially as you would have to keep track of simultaneous red and blue counts and you can't be certain what the color of the cards after the next one will be.

However, with automatic shuffle, it may depend on how long it takes for a dealt card to make its way back into the shuffler. My gut answer is, "Not without either going to great pains to maintain counts." You could try counting just the blue cards, although the problem is the same as before - you have no idea what the backs of the cards after the next one out, if that, are going to be.
aceside
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June 10th, 2023 at 1:18:11 PM permalink
Thank you for sharing your thoughts. Specifically, it was a continuous shuffling machine, One2Six, which exposed the back of the card to be dealt. I noticed some veteran players strived to seat in the first base, and this made me think that it had something to do with the back color information. Let us just assume that all dealt cards were fed back into the shuffler immediately after the first round of playing.

Here is what I think about this. To play this game, we need to adjust the basic strategy, especially on the hand of 16vs10. If all the exposed cards are of the same back color, we hit if the next card has the same back color, but we stand if the next card has a different back color.
Wizard
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June 12th, 2023 at 2:42:27 PM permalink
Quote: aceside

I recently played a 6-deck automatic-shuffle blackjack game. Everything was normal except one thing, that is, 3 of the 6 decks of cards had a red back and the remaining 3 decks of cards had a blue back. Can a player gain some EV edge from this game by taking into account the card back color information?
link to original post



A small benefit could be gained by keeping separate blue and red counts. I would only do this with a confederate, one counting each color, and communicating the information. However, I don't think the gain in EV would be enough to warrant the bother. I think more money could be made with both counters at separate tables.

By the way, I think you should have made a separate thread for this. I like to keep this thread on just fun math and logic puzzles. I'll split this discussion off it engenders more than 2 or 3 more posts.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Wizard
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June 12th, 2023 at 2:44:49 PM permalink
I apologize if I've asked this before.

A prisoner is presented with three closed doors, labeled A, B, and C. The warden explains that behind two doors is freedom and behind one is immediate death. The prisoner may point to one door and ask one yes/no question of the warden.

If the prisoner points towards a door leading to freedom, the warden will answer truthfully. If the prisoner points towards a door leading to death, the warden will answer yes or no randomly.

What should the prisoner do?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
avianrandy
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June 12th, 2023 at 3:24:00 PM permalink
Quote: Wizard

I apologize if I've asked this before.

A prisoner is presented with three closed doors, labeled A, B, and C. The warden explains that behind two doors is freedom and behind one is immediate death. The prisoner may point to one door and ask one yes/no question of the warden.

If the prisoner points towards a door leading to freedom, the warden will answer truthfully. If the prisoner points towards a door leading to death, the warden will answer yes or no randomly.

What should the prisoner do?
link to original post

I thought this question sounded familiar so I used the search function here. You asked it March of last year. It is on page 262 of this thread. It is still a good question though
Wizard
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June 12th, 2023 at 9:27:29 PM permalink
Quote: avianrandy

I thought this question sounded familiar so I used the search function here. You asked it March of last year. It is on page 262 of this thread. It is still a good question though
link to original post



Good memory!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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